A capacitor is constructed from two square metal plates of side length L and sep
ID: 2026805 • Letter: A
Question
A capacitor is constructed from two square metal plates of side length L and separated by a distance d (Figure 6.23). One half of the space between the plates (top to bottom) is filled with polystyrene (k=2.56), and the other half is filled with neoprene rubber (k=6.7). Calculate the capacitance of the device, taking L=2cm and d=0.75mm. (Hint: The capacitor can be considered as two capacitors connected in parallel.)
Explanation / Answer
Solution: side lenght of the plates , L=2 cm(10-2 m/1 cm) =0.02 m Let, the area of the each square plate, A = (Ld) / 2 side lenght of the plates , L=2 cm(10-2 m/1 cm) =0.02 m =(0.02 m)(0.75x10-3 m)/2 =0.0075 x10-3 m2 distance between the plates, d= 0.75 mm(10-3 m/1mm) =0.75x10-3 m di-electric constant of polystyrene (k 1=2.56) fills in the upper half of the capacitor di-electric constant of polystyrene (k 2=6.70) fills in the bottom half of the capacitor permitivity of the free- space (or) air medium is,o =8.54x10-12 Nm2/C2 let, capacitence of the capacitor in upper half of the capacitence be C1 capacitence of the capacitor in upper half of the capacitence be C2 C be the resultant capacitence of C1 and C2 _____________________________________________________________________ but ,capacitence of the parallel plate capacitor C =K A o /d ( General formula) since, these capacitences considered to connect in parallel Resultant capacitnce of these, C = C1 + C2 = k1 Ao / d + k2 Ao / d = [Ao / d ](K1 +K2 ) = [(0.0075 x10-3 m2)(8.54x10-12 Nm2/C2)/(0.75x10-3 m)](2.56+6.70) = 0.0480375x10-12 (9.26) F = 0.4448x10-12 F (or) = 0.448 pF = k1 Ao / d + k2 Ao / d = [Ao / d ](K1 +K2 ) = [(0.0075 x10-3 m2)(8.54x10-12 Nm2/C2)/(0.75x10-3 m)](2.56+6.70) = 0.0480375x10-12 (9.26) F = 0.4448x10-12 F (or) = 0.448 pF (or) = 0.448 pFRelated Questions
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